我想知道我如何可以初始化一个数组(或列表)，尚未与值填充，有一个定义的大小。
例如在C语言中:
int x[5]; /* declared without adding elements*/
在Python中如何做到这一点?
当前回答
为什么这些问题没有得到显而易见的答案呢?
a = numpy.empty(n, dtype=object)
这将创建一个长度为n的数组，用于存储对象。它不能被调整大小或追加。特别是，它不会因为填充长度而浪费空间。这是Python中与Java相同的部分
Object[] a = new Object[n];
如果你真的对性能和空间感兴趣，并且知道你的数组只存储某些数值类型，那么你可以将dtype参数更改为其他一些值，比如int。然后numpy将这些元素直接打包到数组中，而不是使数组引用int对象。
其他回答
为什么这些问题没有得到显而易见的答案呢?
a = numpy.empty(n, dtype=object)
这将创建一个长度为n的数组，用于存储对象。它不能被调整大小或追加。特别是，它不会因为填充长度而浪费空间。这是Python中与Java相同的部分
Object[] a = new Object[n];
如果你真的对性能和空间感兴趣，并且知道你的数组只存储某些数值类型，那么你可以将dtype参数更改为其他一些值，比如int。然后numpy将这些元素直接打包到数组中，而不是使数组引用int对象。
我发现很容易做的一件事是我设置一个数组 例如，我喜欢的大小的空字符串
代码:
import numpy as np
x= np.zeros(5,str)
print x
输出:
['' '' '' '' '']
希望这对你有帮助:)
这样做:
>>> d = [ [ None for y in range( 2 ) ] for x in range( 2 ) ]
>>> d
[[None, None], [None, None]]
>>> d[0][0] = 1
>>> d
[[1, None], [None, None]]
其他的解决方案会导致这样的问题:
>>> d = [ [ None ] * 2 ] * 2
>>> d
[[None, None], [None, None]]
>>> d[0][0] = 1
>>> d
[[1, None], [1, None]]
您可以尝试使用Descriptor来限制大小
class fixedSizeArray(object):
def __init__(self, arraySize=5):
self.arraySize = arraySize
self.array = [None] * self.arraySize
def __repr__(self):
return str(self.array)
def __get__(self, instance, owner):
return self.array
def append(self, index=None, value=None):
print "Append Operation cannot be performed on fixed size array"
return
def insert(self, index=None, value=None):
if not index and index - 1 not in xrange(self.arraySize):
print 'invalid Index or Array Size Exceeded'
return
try:
self.array[index] = value
except:
print 'This is Fixed Size Array: Please Use the available Indices'
arr = fixedSizeArray(5)
print arr
arr.append(100)
print arr
arr.insert(1, 200)
print arr
arr.insert(5, 300)
print arr
输出:
[None, None, None, None, None]
Append Operation cannot be performed on fixed size array
[None, None, None, None, None]
[None, 200, None, None, None]
This is Fixed Size Array: Please Use the available Indices
[None, 200, None, None, None]
>>> n = 5 #length of list
>>> list = [None] * n #populate list, length n with n entries "None"
>>> print(list)
[None, None, None, None, None]
>>> list.append(1) #append 1 to right side of list
>>> list = list[-n:] #redefine list as the last n elements of list
>>> print(list)
[None, None, None, None, 1]
>>> list.append(1) #append 1 to right side of list
>>> list = list[-n:] #redefine list as the last n elements of list
>>> print(list)
[None, None, None, 1, 1]
>>> list.append(1) #append 1 to right side of list
>>> list = list[-n:] #redefine list as the last n elements of list
>>> print(list)
[None, None, 1, 1, 1]
或者一开始就什么都没有:
>>> n = 5 #length of list
>>> list = [] # create list
>>> print(list)
[]
>>> list.append(1) #append 1 to right side of list
>>> list = list[-n:] #redefine list as the last n elements of list
>>> print(list)
[1]
在append的第4次迭代中:
>>> list.append(1) #append 1 to right side of list
>>> list = list[-n:] #redefine list as the last n elements of list
>>> print(list)
[1,1,1,1]
5及所有后续:
>>> list.append(1) #append 1 to right side of list
>>> list = list[-n:] #redefine list as the last n elements of list
>>> print(list)
[1,1,1,1,1]
